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Suppose we have a generic Lagrangian density, for example: $$\mathcal{L} = \alpha A_{\mu\nu}A^{\mu\nu} + \beta B_{\mu}f_\nu(p^2) A^{\mu\nu} + \gamma B_\mu\partial^\mu h$$ where $A_{\mu\nu}$,$B_\mu$ and $h$ are generic fields, $f_\nu(p^2)$ a generic function and $\alpha$,$\beta$,$\gamma$ some real parameters.

I have doubts about the calculation of the propagators for the system of fields $\{A_{\mu\nu},B_\mu,h\}$, correct me if I'm wrong.

I know that the propagator of a field is the green function for its equation and in general to calculate them I put the equations of the fields in Fourier space inside a matrix and look for the inverse.

My question is if, just by looking at the Lagrangian density, I can can say something about the propagators. For example, is it correct to say that since there are no quadratic terms in $B_\mu$ and in $h$, their propagators are null? And since there are no coupling terms between $A_{\mu\nu}$ and $h$ the propagator $$\langle0|T[A_{\mu\nu}^\dagger(x)h(0)]|0\rangle$$ is also null?

Qmechanic
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Fred
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  • Hi and welcome to PSE. I have never seen a propagator as a two point function between one type of field and another. Why do you think that there should exist such a propagator? What would the physical meaning of something like that be? Also, keep in mind that the propagator is derived from the terms in the Lagrangian that are quadratic wrt a specific type of field and they also contain derivatives of that field. – schris38 Sep 13 '22 at 09:49
  • Yes, I agree that the propagators between different types of fields are not part of Feynamn diagrams but one could calculate them a priori. – Fred Sep 13 '22 at 10:23

1 Answers1

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  1. Formally speaking, if the Hessian matrix of the quadratic action is non-degenerate, the free propagator is given by the inverse matrix, cf. e.g. my related Phys.SE answer here.

  2. Concerning OP's last question: If a matrix element is zero, the corresponding inverse matrix element might not necessarily be zero.

Qmechanic
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  • Thanks for the reply. So one way to calculate propagators (even those between different types of fields like the one I wrote) is to find the inverse of the matrix. But to reduce the calculations, is it correct to observe that since there are no quadratic terms in the fields B and h, the latter are not fields of matter and therefore their propagator is necessarily null? – Fred Sep 13 '22 at 10:23
  • I didn't check your example, but not generically. – Qmechanic Sep 13 '22 at 10:34